Chess in Flatland
In Flatland (page 255) the world is a plane and its inhabitants are geometric shapes. Flatlanders play their own versions of Spaceland games, and one of them is chess. The Flatland chessboard is eight cells long, and each player has three pieces: king, knight and rook, starting in the position shown.
Start of a game of Flatland chess.
The rules resemble those of Spaceland chess, bearing in mind the limitations of Flatland geometry. All three pieces can move to the left or to the right, if a suitable space is available. All moves must end either on an empty cell, or on a cell occupied by an enemy piece, which is then removed from the board - ‘taken’.
• A king (the piece with a cross on top) moves only one cell at a time, and cannot move into ‘check’ - a cell that is already threatened by the enemy.
• A knight (horse shape) moves by jumping over an adjacent cell, which may be empty or occupied, and landing on the one on its far side. So it ends up two cells away from where it started.
• A rook (castle shape) can move across any number of unoccupied cells.
If a player has no legal move available, the game is stalemate, and is a draw. If a player can threaten the opposing king, and the king cannot escape, that’s checkmate, and the game is won.
If White plays first, and both players adopt perfect strategy, who will win?
Answer on page 337
The Infinite Lottery
The Infinite Lottery involves infinitely many bags: one numbered 1, one numbered 2, one numbered 3, one numbered 4, and so on. Each bag contains infinitely many lottery balls with the corresponding number.
You are supplied with a large box. You may place any number of balls you like into this box, chosen from any of the bags. There is just one condition: the total number must be finite.
Now you are required to change the balls in the box. You must remove and discard one, and replace it by as many balls as you like that bear smaller numbers. For instance, if you discard a ball with the number 100 on it, you can add to the box 10 million balls with the number 99, 17 billion with the number 98, and so on. There is thus no upper limit to the number of balls that can replace that solitary number-100 ball.
You must keep doing this. At each stage you may replace the discarded ball by whichever combination of balls you wish, provided you make their number finite and make sure that they bear smaller numbers than the ball you discarded. If you remove a ball marked 1, you cannot replace it because there are no balls with smaller numbers on them.
If eventually you run out of balls and empty the box, you lose. If you keep removing balls for ever - that is, if you never run out of balls - then you win.
But can you win the infinite lottery? If so, how?
Answer on page 337
Ships That Pass ...
In the days when people crossed the Atlantic in passenger liners, a ship left London every day at 4.00 p.m. bound for New York, arriving exactly 7 days later.
Every day at the same instant (11.00 a.m. because of the time difference) a ship left New York bound for London, arriving exactly 7 days later.
All ships followed the same route, deviating slightly to avoid collisions when they met.
How many ships from London does each ship sailing from New York encounter during its transatlantic voyage, not counting any that arrive at the dock just as they leave, or leave the dock just as they arrive?
Answer on page 339
The Largest Number Is Forty-Two
Mathematicians often use a technique called proof by contradiction. The idea is that to prove some statement true, you begin by assuming it to be false, and go on to derive various logical consequences. If any of these consequences leads to a logical impossibility - a contradiction - then your assumption that the statement is false cannot be correct. Therefore the statement is true.
You may have come across this by the name used in Euclid, which in Latin translation is reductio ad absurdum - reduction to the absurd.
For example, to prove that pigs don’t have wings, you first assume that they do, and deduce that pigs can fly. But we know that they can’t, so this is a logical impossibility. Therefore it is false that pigs don’t have wings, so they do.
Got that?
I will now use proof by contradiction to show that the largest whole number is 42.
Let n be the largest whole number, and suppose for a contradiction that n is not 42. Then n > 42, so (n - 42)3 >0, which expands to give
n3 - 126n2 + 5,292n - 74,088 > 0
Adding n to each side,
n3 - 126n2 + 5,293n - 74,088 > n
But the left-hand side is a whole number. Since it is greater than n, which we are assuming to be the largest whole number, we have derived a contradiction.
Therefore it is false that the largest whole number is not 42. So it is 42!
Clearly something is wrong here - but what?
Answer on page 339
A Future History of Mathematics
2087 Fermat’s Lost Theorem is found again on the back of an old hymn-sheet in the Vatican secret archives.
2132 A general definition of ‘life’ is formulated at the Intercontinental Congress of Biomathematicians.
2133 Kashin and Chypsz prove that life cannot exist.
2156 Cheesburger and Fries prove that at least one of Euler’s constant, the Feigenbaum number, and the fractal dimension of the universe is irrational.
2222 The consistency of mathematics is established - it is that of cold sago pudding.
2237 Marqès and Spinoza prove that the undecidability of the undecidability of the undecidability of the undecidability of the P=NP? problem is undecidable.
2238 Pyotr-Jane Dumczyk disproves the Riemann hypothesis by showing that there exist at least 42 zeros σ + it of the zeta function with σ ≠ and t < exp exp exp exp exp ((πe +eπ ) log 42).
2240 Fermat’s Lost Theorem is lost again.
2241 Sausage conjecture proved in all dimensions except 5, with the possible exception of the 14- dimensional case, where the proof remains controversial since it seems too easy.
2299 Contact is made with aliens from Grumpius, whose mathematics includes a complete classification of all possible topologies for turbulent flows, but has been stuck for the past five galactic revolutions because of an inability to solve the 1 + 1 = ? problem.
2299 The solution of the 1 + 1 = ? problem by Martha Snodgrass, a six-year-old schoolgirl from Woking, ushers in a new age of Terran-Grumpian cooperation.
2300 Formulation of Dilbert’s 744 problems at the Interstellar Congress of Mathematicians.
2301 Grumpians depart, citing the start of the cricket season.
2408 Riculus Fergle uses Grumpian orthocalculus to show that all of Dilbert’s problems are equivalent to each other, thereby reducing the whole of mathematics to a single short formula.*
2417 The DNA-superstring computer Vast Intellect fails the Turing Test on a technicality but declares itself intelligent anyway.
2417 Vaster Intellect invents the technique of human-assisted proof, and uses it to prove Fergle’s Final Formula, with the Dilbert problems as corollaries.
2417 Even Vaster Intellect discovers inconsistencies in the operating system of the human brain, and all human-assisted proofs are declared invalid.
7999 Grunt Snortsen invents counting on his toes; Reign of the Machines comes to an abrupt end.†
* The famous € ♋♍☾42 . Plus a constant.
† Snortsen had lost a toe in an encounter with a berserk cashregister.
11,868 The Rediscovery of Mathematics, now in base 9.
0 Reformation of the calendar.
1302* Fergle’s Final Formula is proved, correctly this time, and mathematics stops.
1302† Diculus Snergle asks what would happen if you allowed the arbitrary constant in Fergle’s Final Formula to be a variable, and mathematics starts up again.
50 51
Professor Stewart’s Superlative Storehouse of Sneaky Solutions
and Stimulating Supplements
Wherein the perspicacious or perplexed reader may procure answers to those questions that are presently known to possess answers ... together with such gratuitous facts and fancies as may facilitate their further delectation and enlightenment.
Calculator Curiosity 1
(8×8) + 13 = 77
(8×88) + 13 = 717
(8×888) + 13 = 7117
(8×8888) + 13 = 71117
(8×88888) + 13 = 711117
(8×888888) + 13 = 7111117
(8×8888888) + 13 = 71111117
(8×88888888) + 13 = 711111117
Year Turned Upside Down
Past: 1961; Future: 6009.
If you insist on allowing a squiggle on the 7, amend these to 2007 and 2117.
Sixteen Matches
Move these two.
Swallowing Elephants
The deduction is false.
Suppose, for the sake of argument, that elephants are easy to swallow. Then the third statement in the puzzle tells us that elephants eat honey. The second then tells us that elephants can play the bagpipes. On the other hand, the first statement tells us that elephants wear pink trousers, in which case the fourth statement tells us that elephants can’t play the bagpipes. So we get a logical contradiction. The only way out is if elephants are not easy to swallow.
There’s a systematic method for answering such questions. First, turn everything into symbols. Let
E be the statement: ‘Is an elephant.’
H be the statement: ‘Eats honey.’
S be the statement: ‘Is easy to swallow.’
P be the statement: ‘Wears pink trousers.’
B be the statement: ‘Can play the bagpipes.’
We use the logical symbols
⇒ meaning ‘implies’
¬ meaning ‘not’.
Then the first four statements read:
E ⇒ P
H ⇒ B
S ⇒ H
P ⇒ ¬ B
We need two of the mathematical laws of logic:
X ⇒ Y is the same as ¬ Y ⇒ ¬ X
If X ⇒ Y ⇒ Z, then X ⇒ Z
Using these, we can rewrite the conditions as:
E ⇒ P ⇒ ¬ B ⇒ ¬ H ⇒ ¬ S
so E ⇒ ¬ S. That is, elephants are not easy to swallow.
This list of attributes suggests yet another way to get the answer: think about an elephant (E) that (P) wears pink trousers, (¬ B) does not play the bagpipes, (¬ H) does not eat honey, and (¬ S) is not easy to swallow. Then all four statemnts in the puzzle are true, but ‘elephants are easy to swallow’ is false.
Magic Circle
These or their rotations and reflections.
Press-the-Digit-ation
The explanation of Whodunni’s calculator trick uses a bit of algebra.
Suppose you live in house number x, were born in year y, and have had z birthdays so far this year, which is either 0 or 1, depending on dates. Then successive steps in the trick go like this:
• Enter your house number: x
• Double it: 2x
• Add 42: 2x + 42
• Multiply by 50: 50(2x + 42) = 100x + 2100
• Subtract the year of your birth: 100x + 2100 - y
• Subtract 50: 100x + 2050 - y
• Add the number of birthdays you have had this year: 100x + 2050 - y + z
• Subtract 42: 100x + 2008 - y + z
If we’re doing the trick in 2009, then 2008 - y is one less than the number of years that have passed since your birth year. Adding the number of birthdays you’ve had this year leaves it that way if you haven’t had one yet, but adds 1 if you have. The result is always your age. (Think about it. If you were born one year ago but haven’t had a birthday yet, your age is 0. After your birthday, it’s 1.)
So the final result is 100x + your age. So provided you are aged between 1 and 99, the last two digits will be your age (written as 01- 09 if your age is 1-9). Removing those and dividing by 100, which is the same as looking at the remaining digits, gives x - your house number.
If you’re over 99, then the final two digits can’t be your age. There will be an extra digit (which barring medical miracles will be 1). So your age will be 1 followed by the final two digits. And your house number will be the rest of the digits except those two, minus one.
If you’re age 0, the trick still works provided you count the day of your birth as a birth-day. Your zeroth, in fact. But usually we don’t do that, which is why I excluded age 0.
To modify the trick for any other year, say 2009 + a, just change the final step to ‘subtract 42 - a’. So in 2010 subtract 41, in 2011 subtract 40, and so on. If you’re reading this after 2051, make that ‘add a - 42’. It’s the same thing, but it will sound more sensible.
Secrets of the Abacus
To subtract (say) a 3-digit number [x][y][z], which is really 100x + 10y + z, we have to form the complement [10 - x][10 - y] [10 - z], which is really 100(10 - x) + 10(10 - y) + (10 - z). This is equal to 1000 - 100x + 100 - 10y + 10 - z, or 1110 - (100x + 10y + z). So adding the complement is the same as subtracting the original number, but adding 1110. To get rid of it, subtract 1 from positions 4, 3, 2, but not 1.
Redbeard’s Treasure
Redbeard will locate the lost loot 128 paces north of the rock.
At each step the piratical finger can move either left or right - two choices. So the number of routes down the diagram doubles for each extra row. There are 8 rows, and only one T to start from, so the number of routes is 1×2×2×2×2×2×2×2 = 128.
If we replace each letter by the number of routes that lead to it, we get a famous mathematical gadget, Pascal’s triangle:
Here each number is the sum of the two above it to left and right, except down the sides where they’re all 1’s. If you add up the rows, you get the powers of two: 1, 2, 4, 8, 16, 32, 64, 128. So this is another - closely related - way to get the same answer.
Stars and Snips
Fold the paper in half (say along the vertical line in my picture) and then fold it alternately up and down along the other lines to make a zigzag shape, the way a fan folds. Then cut along a suitable slanting line - and unfold. I’ve drawn the ghost of the star to show how it relates to the folds.
Fold . . .
... and cut.
Yes, you can make a six-pointed star in a similar way. If anything, it’s easier: first fold the paper in quarters, then fold the result along two lines trisecting the right-angled corner. As for the five-pointed star, you have to snip at the correct angle. That’s committees for you.
The Collatz-Syracuse-Ulam Problem
The cycles that appear with zero or negative numbers are:
• 0→0
• -1 → -2 → -1
• -5 → -14 → -7 → -20 → -10 → -5
• -17 → -50 → -25 → -74 → -37 → -110 → -55 → -164 → -82 → -41 → -122 → -61 → -182 → -91 → -272 → -136 → -68 → -34 → -17
The Jeweller’s Dilemma
The lengths contain 8, 7, 6, 6, 5, 5, 5, 4 and 3 links. Instead of breaking up one link on each chain, we could break all 8 links in the 8-link piece, and use these to join the other eight pieces together: total cost £24. But there’s a cheaper way. Break the pieces of lengths 4 and 3 into separate links, and use these to join the seven other pieces. The total cost is now £21.
What Seamus Didn’t Know
No, it doesn’t wave its paws madly and exploit air resistance to create a force, like the wings of a bird do. Instead, the cat manages to change its orientation without causing any change to its angular momentum at any time.
• Initial position: cat upside down, stationary, angular momentum zero.
• Final position: cat right side up, stationary, angular momentum zero.
No contradiction there, but of course there’s the bit in between, when the cat starts to rotate. Except - it doesn’t. Rotate, that is. A cat is not a rigid body.52 In 1894, the French doctor Étienne J
ules Marey took a series of photographs of a falling cat.
Marey’s cat experiment.
The secret was then revealed. Because a cat is not a rigid body, it does not have to rotate its entire body simultaneously. Here is the cat’s recipe for turning over, while maintaining zero angular momentum throughout:
• Pull in your front legs and spread out your back legs.
• Twist your front half quickly one way, and your back half slowly the other way. Your two halves move with opposite angular momentum, so the total remains zero.
• Spread out your front legs and pull in your back legs.
• Turn your back half quickly to align with the front half, while your front half turns slowly backward. Again your two halves move with opposite angular momentum, so the total remains zero.
• Your tail can also move, and usually does, assisting the process by providing a useful reservoir of spare angular momentum.
Modern photo of falling cat.
Lincoln’s Dog
Well, the dog may have lost its tail, or some legs, or it might be a mutant with six legs and five tails ... Quibbles aside, this is a good question to distinguish mathematicians from politicians. Lincoln asked his question in the context of slavery, putting it to a political opponent who maintained that slavery was a form of protection, trying to imply that it was benign. Lincoln’s answer was: ‘It has four legs - calling a tail a leg doesn’t make it a leg.’ By which he meant that calling slavery protection didn’t make it protection, which in that context is fair enough. Barack Obama’s famous ‘lipstick on a pig’ remark made the same point, though his opponents chose to interpret it as an insult to Sarah Palin.53
Ian Stewart Page 22